MATH 162, SHEET 8: the REAL NUMBERS 8A. Construction of The

MATH 162, SHEET 8: THE REAL NUMBERS

This sheet is concerned with proving that the continuum R is an ordered ﬁeld. Addition and multiplication on R are deﬁned in terms of addition and multiplication on Q, so we will use ⊕ and ⊗ for addition and multiplication of real numbers, to make sure that there is no confusion with + and · on Q.

8A. Construction of the real numbers via cuts, Addition and Order

Deﬁnition 8.1. A subset A of Q is said to be a cut (or Dedekind cut) if it satisﬁes the following:

(a) A 6= Ø and A 6= Q (b) If r ∈ A and s ∈ Q satisfies s < r, then s ∈ A (c) If r ∈ A then there is some s ∈ Q with s > r, s ∈ A. We denote the collection of all cuts by R. Definition 8.2. We define ⊕ on R as follows. Let A, B ∈ R be Dedekind cuts. Define A ⊕ B = {a + b | a ∈ A and b ∈ B} 0 = {x ∈ Q | x < 0} 1 = {x ∈ Q | x < 1}. Exercise 8.3. (a) Prove that A ⊕ B, 0, and 1 are all Dedekind cuts.

(b) Prove that {x ∈ Q | x ≤ 0} is not a Dedekind cut. (c) Prove that {x ∈ Q | x < 0} ∪ {x ∈ Q | x2 < 2} is a Dedekind cut. Hint: In Exercise 4.20 last quarter we showed that if x ∈ Q, x ≥ 0, and x2 < 2, then there is some δ ∈ Q, δ > 0 such that (x + δ)2 < 2. Exercise 8.4. (Homework) Prove that if A ∈ R then A = 0 ⊕ A. Deﬁnition 8.5. If A, B ∈ R, we say that A < B if A is a proper subset of B, i.e. A ⊂ B but A 6= B. Exercise 8.6. Show that < satisﬁes

(a) If A, B ∈ R then exactly one of the following holds: A < B, A = B, B < A.

(trichotomy)

1 (b) If A, B, C ∈ R with A < B and B < C then A < C. (transitivity)

(c) If A, B, C ∈ R with A < B then A ⊕ C < B ⊕ C. (additivity) Note: to prove part c) you may assume Theorem 8.11 and associativity of addition.

Exercise 8.7. (Homework) Show that if A ∈ R and 0 < A, then 0 ∈ A.

Lemma 8.8. Let A ∈ R. Then deﬁne

−A = {r ∈ Q | −r 6∈ A but − r is not the least element of Q \ A}. Then

(a) −A = {r ∈ Q | ∃ > 0, ∈ Q such that − r − 6∈ A}

(b) −A ∈ R (c) −(−A) = A

Exercise 8.9. (Homework) Show that if p ∈ Q and A = {x ∈ Q | x < p}, then −A = {x ∈ Q | x < −p}.

Lemma 8.10. If A ∈ R, r ∈ Q, r > 0, then there exist s ∈ A and t ∈ Q \ A such that t − s = r.

Theorem 8.11. If A ∈ R then A ⊕ (−A) = 0. Corollary 8.12. A < 0 ⇐⇒ 0 < −A.

8B. Multiplication, the real numbers form an ordered ﬁeld

Deﬁnition 8.13. For A, B ∈ R, 0 < A, 0 < B, we deﬁne

A ⊗ B = {ab | a ∈ A, b ∈ B, a > 0, b > 0} ∪ {x ∈ Q | x ≤ 0}. If A = 0 or B = 0 we deﬁne A ⊗ B = 0. If A < 0 but 0 < B we replace A with −A and use the deﬁnition of multiplication of positive elements. Hence, in this case,

A ⊗ B = −[(−A) ⊗ B].

Similarly, if 0 < A but B < 0, then

A ⊗ B = −[A ⊗ (−B)] and if A < 0, B < 0 then A ⊗ B = [(−A) ⊗ (−B)]

Exercise 8.14. (a) Show that if A, B ∈ R, then A ⊗ B ∈ R.

2 (b) Show that if A, B ∈ R, 0 < A, 0 < B =⇒ 0 < A ⊗ B.

Definition 8.15. If A ∈ R, 0 < A, then define 1 1 A−1 = {r ∈ | r ≤ 0}∪{r ∈ | r > 0 and 6∈ A, but is not the smallest element of \A}. Q Q r r Q For A < 0 we have 0 < −A and so (−A)−1 can be defined as above. Then, for A < 0, we define A−1 = −[(−A)−1].

Lemma 8.16. Let A ∈ R. Then, (a) If 0 < A, 1 A−1 = {r ∈ | r ≤ 0} ∪ {r ∈ | r > 0 and ∃ > 0, ∈ such that 6∈ A}. Q Q Q r +

(b) If A 6= 0,A ∈ R, then A−1 ∈ R.

Exercise 8.17. Show that if p ∈ Q with p 6= 0 and A = {x ∈ Q | x < p}, then A−1 = {x ∈ 1 Q | x < p }.

Lemma 8.18. If A ∈ R, A > 0, r ∈ Q, r > 1, then there exist s ∈ A, t ∈ Q \ A such that t s = r. Theorem 8.19. If A ∈ R,A 6= 0, then A ⊗ A−1 = 1.

Theorem 8.20. R is an ordered ﬁeld.

8C. The Archimedean property of the real numbers

For every rational number q ∈ Q, deﬁne the corresponding real number as the Dedekind cut

i(q) = {x ∈ Q | x < q}.

For example, 0 = i(0). This gives a well-defined injective function i : Q −→ R. We identify Q with its image i(Q) ⊂ R so that the rational numbers Q are a subset of the real numbers R. (Hence N and Z are also subsets of R.) Note that in this case, this is the same map i that you identified in Theorem 7.27, so that in fact, Q is a subfield of R.

Exercise 8.21. Show that R satisﬁes Axioms 1-3.

Lemma 8.22. A nonempty subset of R that is bounded above has a supremum.

Exercise 8.23. Show that R satisﬁes Axiom 4.

3 Lemma 8.24. N is an unbounded subset of R. Theorem 8.25. (The Archimedean Property) Let A ∈ R be a positive real number. Then 1 there exist nonzero natural numbers m, n ∈ N such that n < A < m. Corollary 8.26. If A ∈ R is a real number, then there is an integer n such that n − 1 ≤ A < n. Deﬁnition 8.27. Consider a subset X ⊂ C of the continuum. We say that X is dense in C if every p ∈ C is a limit point of X. Lemma 8.28. A subset X ⊂ C is dense in C if, and only if, X = C. Lemma 8.29. A subset X ⊂ C is dense in C if, and only if, for every non-empty open set U ⊂ C, we have U ∩ X 6= Ø. Lemma 8.30. Given A, B ∈ R with A < B, there exists p ∈ Q such that A < p < B. Theorem 8.31. Q is dense in R.

8D. The continuum is isomorphic to the real numbers Axiom 5. The continuum contains a countable dense subset. Exercise 8.32. Let K ⊂ C be a countable dense subset of C. Construct an order-preserving bijection f : Q −→ K. (A map f : Q −→ K is order-preserving if r < s in Q =⇒ f(r)

A < B =⇒ F (A)

Axioms 1-5 completely characterize the continuum in the sense that there is only one model for them, namely the real numbers. The main consequence for us is that we will no longer refer to C as anything but R. In addition, now that we have constructed R and proved the fundamental facts about it, we will forget about Dedekind cuts and think of elements of R as numbers. From now on, we will agree to use lower-case letters like x for real numbers and to write + and · for ⊕ and ⊗ like people usually do.